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A student has a sum of money. The first day he spent 3/10 and 2 dolars from them. The next day 2(5 from the rest and another 5 dolars, and the third day the remaining 40 dolars. Find out how much the student had and how much spent every day.

User Kishan Oza
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Final answer:

Working backward from the third day, when the student had $40 left, calculations reveal that the student spent $75 on the second day, which was $5 plus 2/5 of what was left after Day 1. On Day 1, the student spent $2 plus 3/10 of their original amount, which is found to be $110 through backward calculations.

Step-by-step explanation:

To find out how much money the student originally had, we need to work backwards from the third day. Let's denote the original sum as x dollars.

Day 3

The student has $40 left. This means that after spending on Day 2, they were left with $40.

Day 2

The student spends $5 plus 2/5 of the rest. Let's call the remaining money after Day 1 y. So, the equation for Day 2's spending is $5 + 2/5(y). After spending, they are left with $40, so we can write:

y - (5 + 2/5(y)) = 40

Now let's solve for y:

y - 5 - 2/5(y) = 40

5/5(y) - 2/5(y) = 40 + 5

3/5(y) = 45

y = 45 / (3/5)

y = 45 * 5/3

y = 75

Thus, the student had $75 remaining at the beginning of Day 2.

Day 1

The student spends $2 plus 3/10 of the original sum. So, the money left after Day 1 spending would be:

x - (2 + 3/10(x)) = y

Substitute $75 for y:

x - 2 - 3/10(x) = 75

10/10(x) - 3/10(x) = 75 + 2

7/10(x) = 77

x = 77 / (7/10)

x = 77 * 10/7

x = 110

Thus, the student originally had $110.

User Cmidi
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